Cubic Gauss Sums

We compute the cubic subfield of \(\mathbb Q(\zeta_p)\) when \(p\equiv1\pmod3\).

Theorem 1. Fix a prime \(p\) such that \(p\equiv1\pmod3\).

1. There is a unique pair of integers \((A,B)\) for which \(4p=A^2+27B^2\) and \(A\equiv2\pmod3\) and \(B>0\).
2. There is a unique cubic subfield of \(\mathbb Q(\zeta_p)\), and it is generated by a root of the equation \(x^3-3px+pA\).

Quickly, we remark that one can adjust the sign of \(A\) to instead have \(A\equiv1\pmod3\), which makes the equation at the end \(x^3-3px-pA\).

We will show the two parts more or less independently. Because it is easier, let’s start with the first part. Throughout today, \(\omega\) is a primitive third root of unity.

Lemma 2. Fix a nonzero element \(\alpha\) of \(\mathbb Z[\omega]\) which is coprime to \(3\). Then \(\alpha\) admits a unique associate \(\alpha’\) such that \(\alpha’\equiv1\pmod3\).

Proof. We are only interested in\(\pmod3\) information, so it is enough to show that the composite map \[\mu_6=\mathbb Z[\omega]^\times\to\left(\frac{\mathbb Z[\omega]}{(3)}\right)^\times\] is an isomorphism. (Indeed, given \(\alpha\), we are hunting for a unit \(u\) for which \(u\alpha\equiv1\pmod3\), which is equivalent to having a unique unit \(u\) with \(u\equiv\alpha^{-1}\pmod3\).) To show this, \(\mathbb Z[\omega]/(3)\) is represented by the elements \[\left\{\pm1,\pm\omega,\pm(1+\omega)\right\}\sqcup\{0,\pm(1-\omega)\}.\] The left set is \(\mu_6\), and the elements of the right set share a prime factor with \(3\) because \(3=(1-\omega)(2+\omega)\). \(\blacksquare\)

Proposition 3. Fix a prime \(p\) such that \(p\equiv1\pmod3\). There is a pair of integers \((A,B)\) for which \(4p=A^2+27B^2\) and \(A\equiv2\pmod3\).

Proof. Because \(p\equiv1\pmod3\), we see that \(\left(\frac{-3}p\right)=+1\), so \[\frac{\mathbb Z[\omega]}{(p)}\cong\frac{\mathbb F_p[x]}{\left(x^2+3\right)}\] is not a field, so \(p\) is not a prime element in the principal ideal domain \(\mathbb Z[\omega]\), so we may factor \(p=\pi\overline\pi\) for some \(\pi\). By Lemma 2, we may adjust \(\pi\) by some root of unity so that \(\pi\equiv1\pmod3\).

Now, write \(\pi=a-b\omega\) so that \(\operatorname N_{\mathbb Q(\omega)/\mathbb Q}(\pi)=a^2+ab+b^2\), so \[p=a^2+ab+b^2,\] which rearranges into \[4p=(2a+b)^2+3b^2.\] To finish, note \(\pi\equiv1\pmod3\) requires that \(a\equiv1\pmod3\) and \(b\equiv0\pmod3\), so we may take \((A,B)=(2a+b,b/3)\). \(\blacksquare\)

Remark 4. Conversely, if we write \(4p=A^2+27B^2\) as in Theorem 1, then we may set \(b=3B\) and \(a=(A-b)/2\); note that \(a\) is in fact an integer because \(A\) and \(B\) must have the same parity. Then \(p=a^2+ab+b^2\), so \[\pi=a-b\omega=\frac{A+3B\sqrt{-3}}2\] is a factor of \(p\) with \(\pi\equiv1\pmod3\). By unique prime factorization of \(p\), we see that this \(\pi\) is unique up to conjugation. It follows that the pair \((A,B)\) is unique up to the sign of \(B\).

We now turn to the calculation of the cubic subfield.

Proposition 5. Let \(L/K\) be a finite Galois extension of fields with Galois group \(G\), and choose a subgroup \(H\subseteq G\).

1. For each \(\theta\), we have \[\sum_{h\in H}h\theta\in L^H.\]
2. If \(\theta\in L\) is a normal basis element (i.e., \(\{g\theta:g\in G\}\) is a basis for \(L\) over \(K\)), then \[L^H=K\left(\sum_{h\in H}h\theta\right).\]

Proof. Note that \(g\in G\) fixes \(\sum_{h\in H}h\theta\) if and only if \[\sum_{h\in H}gh\theta=\sum_{h\in H}h\theta.\] For example, if \(g\in H\), then the sum is merely rearranged, so the first part follows. For the second part, having a basis means that the equality in \(L\) is equivalent to an equality \[\{gh\theta:h\in H\}=\{h\theta:h\in H\}\] of sets. In particular, \(g\theta\) is in the right-hand set, so \(g\in H\) follows; conversely, \(g\in H\) implies that the two sets are merely permutations of each other. \(\blacksquare\)

Corollary 6. Fix a prime \(p\equiv1\pmod3\). Then there is a unique cubic subfield of \(\mathbb Q(\zeta_p)\), and it is generated by \[\sum_{i=1}^{p-1}\zeta_p^{i^3}.\]

Proof. Let’s start with the existence and uniqueness of the cubic subfield. The Galois group \(G\) of the extension \(\mathbb Q(\zeta_p)/\mathbb Q\) is the cyclic group \((\mathbb Z/p\mathbb Z)^\times\). The uniqueness of the cubic subfield follows because cyclic groups have unique subgroups of given index (dividing the order). In particular, the unique subgroup of \((\mathbb Z/p\mathbb Z)^\times\) consists of the cubes \[H:=\left\{k^3:k\in(\mathbb Z/p\mathbb Z)^\times\right\}.\] It remains to show that the given element generates \(\mathbb Q(\zeta_p)^H\), which is the purpose of Proposition 5. Indeed, the element \(\zeta_p\) produces a basis \(\{\zeta^i:i\in(\mathbb Z/p\mathbb Z)^\times\}\) for the extension \(\mathbb Q(\zeta_p)/\mathbb Q\), so we see that \[3\sum_{h\in H}h\zeta_p\] generates \(\mathbb Q(\zeta_p)^H\). Because cubing on \((\mathbb Z/p\mathbb Z)^\times\) is \(3\)-to-\(1\) onto \(H\), the result follows. \(\blacksquare\)

It remains to compute the cubic irreducible polynomial of the element defined in Corollary 6. For this, we recall (e.g., from Fourier analysis or representation theory) that it is convenient to replace a sum over an indicator of a subgroup with some character sum. Namely, for a finite abelian group \(A\), the orthogonality relations \[\frac1{\left|A\right|}\sum_{\chi\in\operatorname{Hom}(A,\mathbb C)}\chi\] is the indicator of the identity of \(A\). (We will prove the needed special case of this later.) Because we want to indicate the subgroup of cubes in \((\mathbb Z/p\mathbb Z)^\times\), we want to understand characters of the quotient, which are equivalent to characters \((\mathbb Z/p\mathbb Z)^\times\to\mu_3\).

We will need to do some intricate calculations with our characters, so we take some care in defining our character. The following construction is motivated by Euler’s criterion.

Definition 7. Fix a prime \(\pi\) of \(\mathbb Z[\omega]\) with \(\pi\nmid3\). Then we define \(\chi_\pi\colon(\mathbb Z[\omega]/(\pi))^\times\to\mu_3\) by \[\alpha^{(\operatorname N(\pi)-1)/3}\equiv\chi_\pi(\alpha)\pmod\pi\] for each \(\alpha\). The fraction here is an integer: one can check that \(\operatorname N(\pi)\equiv1\pmod3\) by writing \(\pi=\frac{a+b\sqrt{-3}}2\).

Remark 8. Let’s run some checks on this definition. To see that \(\chi_\pi\) is unique, note that \(\pi\nmid3\) implies that the element of \(\mu_3\) are distinct in \(\mathbb Z[\omega]/(\pi)\): for example, \(1-\omega\) is nonzero because the norm of \(1-\omega\) is \(3\). On the other hand, the element \(\mu_3\subseteq(\mathbb Z[\omega]/(\pi))^\times\) are the unique three solutions to \(\beta^3=1\), so the well-definedness of \(\chi_\pi\) follows by noting \[\left(\alpha^{(\operatorname N(\pi)-1)/3}\right)^3\equiv1\pmod\pi.\] Lastly, we note that \(\chi_\pi\) is multiplicative because \((\cdot)^{(\operatorname N(\pi)-1)/3}\) is multiplicative.

Remark 9. By construction, \(\chi_\pi\) is trivial on the cubes of \((\mathbb Z[\omega]/(\pi))^\times\). On the other hand, let’s check that \(\chi_\pi\) is a nontrivial character. Indeed, because \(\pi\) is prime, we see that \((\mathbb Z[\omega]/(\pi))^\times\) is cyclic, so we may let \(g\) be a generator. Then \[g^{(\operatorname N(\pi)-1)/3}\not\equiv1\pmod{\pi}\] by definition of \(g\), so \(\chi_\pi(g)\ne1\) is forced.

The discussion preceding Definition 7 yields the following lemma.

Definition 10. Fix a prime \(p\). Given a character \(\chi\colon(\mathbb Z/p\mathbb Z)^\times\to\mathbb C^\times\), we define the Gauss sum \[g(\chi)=\sum_{i=1}^{p-1}\chi(i)\zeta^i.\]

Lemma 11. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then \[\sum_{i=1}^{p-1}\zeta_p^{i^3}=g(\chi_\pi)+\overline{g(\chi_\pi)}-1.\] Here, \(\chi_\pi\) is a character on \((\mathbb Z/p\mathbb Z)^\times\) via the inclusion \(\mathbb Z\hookrightarrow\mathbb Z[\omega]\).

Proof. Quickly, note that \(\chi_\pi\) yields a well-defined character on \((\mathbb Z/p\mathbb)^\times\) because the kernel of the ring homomorphism \[\mathbb Z\hookrightarrow\mathbb Z[\omega]\twoheadrightarrow\frac{\mathbb Z[\omega]}{(\pi)}\] is generated by \(\operatorname N_{\mathbb Q(\omega)/\mathbb Q}(\pi)=p\). Observe that this induces an embedding \(\mathbb Z/p\mathbb Z\to\mathbb Z[\omega]/(p)\), which must be an isomorphism because these two rings have the same size.

We now proceed with the proof of the lemma. Let \(H\subseteq(\mathbb Z/p\mathbb Z)^\times\) be the subgroup of cubes. Then on one hand, \[\sum_{i=1}^{p-1}\zeta_p^{i^3}=3\sum_{i\in H}\zeta_p^i,\] as at the end of the proof of Corollary 6. On the other hand, for \(i\in(\mathbb Z/p\mathbb Z)^\times\), we claim that \[1+\chi_\pi(i)+\overline{\chi_\pi(i)}\stackrel?=\begin{cases}3 & \text{if }i\in H, \\0 & \text{if }i\notin H.\end{cases}\] Let’s explain why this claim completes the proof: summing over \(i\) shows that \[\sum_{i=1}^{p-1}\zeta_p^i+g(\chi_\pi)+\overline{g(\chi_\pi)}=3\sum_{i\in H}\zeta_p^i,\] completing the proof of the lemma.

It remains to prove the claim. Surely if \(i\in H\), then \(i\) is a cube, so \(\chi_\pi(i)=1\) follows by definition. For \(i\notin H\), it is enough to show that \(\chi_\pi(i)\ne1\) because \(1+\omega+\overline\omega=0\). But \(\chi_\pi\) is nontrivial on \((\mathbb Z[\omega]/(\pi))^\times\) by Remark 9, so it is nontrivial on \((\mathbb Z/p\mathbb Z)^\times\). We already know that it is trivial on \(H\), so it descends to a nontrivial character on the three-element group \((\mathbb Z/p\mathbb Z)^\times/H\). However, all nontrivial characters on the three-element group are nontrivial on nontrivial elements, so the result follows. \(\blacksquare\)

Lemma 11 tells us that it will be convenient to instead compute the irreducible cubic for \[\sum_{i=0}^{p-1}\zeta_p^{i^3}=g(\chi_\pi)+\overline{g(\chi_\pi)}.\]

Lemma 12. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then the element \[G=\sum_{i=0}^{p-1}\zeta_p^{i^3}\] satisfies the equation \(G^3-3pG-2\operatorname{Re}g(\chi_\pi)^3=0\).

Proof. Set \(g=g(\chi_\pi)\) for brevity. Using Lemma 11, we simply compute \[\begin{aligned}G^3 &= (g+\overline g)^3 \\&= g^3+\overline g^3+3g\overline g(g+\overline g) \\&= 3g\overline gG+2\operatorname{Re}g^3.\end{aligned}\] The result follows after rearrangement as soon as we note that \(g\overline g=p\) as worked out in this post. \(\blacksquare\)

It remains to compute \(g(\chi_\pi)^3\). As evidenced by the statement of Theorem 1, this must be slightly complicated. Let’s start with some partial information.

Lemma 13. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then \[g(\chi_\pi)^3\equiv1\pmod3.\]

Proof. We are trying to show that \(g(\chi_\pi)^3-1\) is an algebraic integer divisible by \(3\). Using the Frobenius automorphism, we see that \[g(\chi_\pi)^3\equiv\sum_{i=1}^{p-1}\chi_\pi(i)^3\zeta_p^{3i}\pmod3.\] However, \(\chi_\pi(i)^3=1\) for all \(i\), so because \(3\) and \(p\) are coprime, we see that \[g(\chi_\pi)^3\equiv\sum_{i=1}^{p-1}\zeta_p^i\pmod3,\] and now the right-hand side is \(-1\). \(\blacksquare\)

It looks like we can use unique prime factorization in \(\mathbb Z[\omega]\) to pin down \(g(\chi_\pi)\) because \(g(\chi_\pi)\overline{g(\chi_\pi)}=p\) and \(g(\chi_\pi)^3\equiv1\pmod3\). However, this does not make sense because \(g(\chi_\pi)\) is not an element of \(\mathbb Z[\omega]\). Luckily, it is not hard to salvage this idea.

Lemma 14. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\). Then \[g(\chi_\pi)^3\in p\mathbb Z[\omega].\]

Proof. This is fairly tricky. Because \(g(\chi_\pi)\overline{g(\chi_\pi)}=p\), it is enough to show that \(g(\chi_\pi)^2\in\overline{g(\chi_\pi)}\mathbb Z[\omega]\), which we will do by a direct calculation. Indeed, \[\begin{aligned}g(\chi_\pi)^2 &= \sum_{i,j=1}^{p-1}\chi_\pi(i)\chi_\pi(j)\zeta_p^{i+j} \\&= \sum_{k=0}^{p-1}\Bigg(\sum_{i+j=k}\chi_\pi(i)\chi_\pi(j)\Bigg)\zeta_p^k \\&= \sum_{i=1}^{p-1}\chi_\pi(i)\chi_\pi(-j)+\sum_{k=1}^{p-1}\Bigg(\sum_{i+j=1}\chi_\pi(i)\chi_\pi(j)\Bigg)\chi_\pi(k)^2\zeta_p^k. \\&= \chi_\pi(-1)\sum_{i=1}^{p-1}\chi_\pi(i)^2+\Bigg(\sum_{i+j=1}\chi_\pi(i)\chi_\pi(j)\Bigg)g\left(\chi_\pi^2\right).\end{aligned}\] Note that \(\chi_\pi^2=\overline{\chi_\pi}\) because \(\chi_\pi\) has codomain \(\mu_3\), so \(g\left(\chi_\pi^2\right)=\chi_\pi(-1)\overline{g(\chi_\pi)}\) follows after a rearrangement. Because \(\chi_\pi\) is nontrivial, it follows that the left sum vanishes, so the result follows because \(\operatorname{im}\chi_\pi\subseteq\mathbb Z[\omega]\). \(\blacksquare\)

Remark 15. In general, given nontrivial Dirichlet characters \(\chi_1,\chi_2\pmod p\) for which \(\chi_1\chi_2\) is nontrivial, the same argument shows that \[\frac{g(\chi_1)g(\chi_2)}{g(\chi_1\chi_2)}=\sum_{i+j=1}\chi_1(i)\chi_2(j).\] The right-hand sum is important enough to deserve its own name: it is called a Jacobi sum.

Proposition 16. Fix a rational prime \(p\equiv1\pmod3\), and factor \(p=\pi\overline\pi\) in \(\mathbb Z[\omega]\) with \(\pi\equiv1\pmod3\). Then \[g(\chi_\pi)^3\in\{-p\pi,-p\overline\pi\}.\]

Proof. By Lemma 14, \(g(\chi_\pi)^3\in p\mathbb Z[\omega]\). Because \(g(\chi_\pi)\overline{g(\chi_\pi)}=p\), we see that the prime factorization \(g(\chi_\pi)^3/p\) is associate to either \(\pi\) or \(\overline\pi\). To finish, note that we can remove the associate ambiguity by Lemma 2 because \(g(\chi_\pi)^3/p\equiv-1\pmod3\) by Lemma 13. \(\blacksquare\)

We are now ready to collect everything we have done in this blog post.

Proof of Theorem 1. The first part of the theorem follows from Proposition 3 and Remark 4. In fact, Remark 4 has shown that \[\pi=\frac{A+3B\sqrt{-3}}2\] is the unique factor of \(p\) in \(\mathbb Z[\omega]\) with \(\operatorname{Im}\pi>0\) and \(\pi\equiv1\pmod3\).

We now turn to the second part of the theorem. By Corollary 6 and Lemma 12, we see that the cubic subfield is generated by an element \(G\) which is the root of the polynomial \[x^3-3px-2\operatorname{Re}g(\chi_\pi)^3.\] However, \(g(\chi_\pi)^3\in\{-p\pi,-p\overline\pi\}\) by Proposition 16, so \(-2\operatorname{Re}g(\chi_\pi)^3=pA\). \(\blacksquare\)

Remark 17. The element \(\sum_{i=1}^{p-1}\zeta_p^{i^3}\) will not generate the ring of integers in the cubic subfield because it is divisible by \(3\). However, \[\frac13\sum_{i=1}^{p-1}\zeta_p^{i^3}\] also may not generate the ring of integers. For example, if \(p=31\), then the discriminant of this element is divisible by \(4\); however, the discriminant of the cubic subfield must divide the discriminant of \(\mathbb Q(\zeta_p)\), and the latter is a power of \(p\). On the other hand, discriminant calculations are enough to verify that this element generates the ring of integers when \(p\in\{7,13,19,37\}\): in these cases, the discriminant of this element is \(p^2\), but \(p\) does ramify in the cubic subfield.

Remark 18. It is possible to determine which of \(p\pi\) or \(p\overline\pi\) equals \(g(\chi_\pi)^3\) by computing the Jacobi sum \[\sum_{i+j=1}\chi_\pi(i)\chi_\pi(j).\] The answer is slightly complicated; it is worked out as Theorem 3.1.3 in the book Gauss Sums and Jacobi Sums by Berndt, Evans, and Williams.